1. Results of a math standard achievement test when analyzed indicated the scores exhibited a normal distribution. The mean was 620 with standard deviation of 52. (a) Some inquired about the fraction of the test takers achieved at 95% confidence scores greater than 800. Determine that fraction, (b) determine for different math test and 95% confidence the range of scores, but only 24 students took this test. The mean was 73, standard deviation of 11.
1a) Let X be the score in math standard achievement test
X follow Normal distribution
to find P(X> 800)= P(Z> (800-620)/52)
= P(Z> 3.46)
= 0.0003
fraction of test takers who scores greater than 800 is 0.0003
b) 95% confidence inerval for true mean
x ̅ ± zc σ⁄√n
= 73 ± 1.96 * 11/√24
= 73 ± 4.4
= (68.6, 77.4)
note: for 95% CI zc = 1.96
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